Lecture Notes in Astrophysical Fluid Dynamics

Transcription

1 Lecture Notes in Astrophysical Fluid Dynamics Mattia Sormani November 19,

2 Contents 1 Hydrodynamics Introductory remarks The state of a fluid The continuity equation The Euler equation, or F = ma The choice of the equation of state Manipulating the fluid equations Writing the equations in different coordinate systems Indecent indices Tables of unit vectors and their derivatives Conservation of energy Conservation of momentum Lagrangian vs Eulerian view Vorticity The vorticity equation Kelvin circulation theorem Steady flow: the Bernoulli s equation Rotating frames Viscosity and thermal conduction The Reynolds number Adding radiative heating and cooling Summary Problems Magnetohydrodynamics Basic equations Magnetic tension Magnetic flux freezing Magnetic field amplification Conservation of momentum Conservation of energy How is energy transferred from fluid to fields if the magnetic force does no work? Summary Problems Hydrostatic equilibrium Polytropic spheres The isothermal sphere

3 3.3 Stability of polytropic and isothermal spheres Problems Spherical steady flows Parker wind Bondi spherical accretion Problems Waves Introduction Sound waves Propagation of arbitrary perturbations Water waves Group velocity Analogy between shallow water theory and gas dynamics MHD waves

4 These are the lecture notes for the course in astrophysical fluid dynamics at Heidelberg university. The course is jointly taught by Simon Glover and myself. These notes are work in progress and I will update them as the course proceeds. It is very likely that they contain many mistakes, for which I assume full responsibility. If you find any mistake, please let me know by sending an at mattia.sormani@uni-heidelberg.de References In these notes we make no attempt at originality and we draw from the following sources. [1] Acheson, D. J., Elementary Fluid Dynamics, (Oxford University Press, 1990) Notes: a very clear text on fluid dynamics. Highly recommended. [2] Landau, L.D., Lisfshitz, E.M. Fluid Mechanics, (Elsevier, 1987) Notes: part of the celebrated Course of Theoretical Physics series. A fantastic book. [3] Feynman, R.P., The Feynman Lectures on Physics Notes: look at Chapter 40 and 41 in Vol II. The lectures are available for free on the web at the following link: [4] Shore, S.N., Astrophysical Hydrodynamics, (Wiley, 2007). Notes: a very comprehensive text with lots of wonderful historical notes. [5] Balbus, S., Lecture Notes, available here afd1415_pdf_18150.pdf and here hydrosun_pdf_66537.pdf Notes: very clear and concise. [6] Clarke, C.J., Carswell, R.F., Astrophysical Fluid Dynamics, (Cambridge University Press, 2007). Notes: a modern textbook on astrophysical fluid dynamics. [7] Binney, J., Tremaine, S., Galactic Dynamics, (Princeton Series in Astrophysics, 2008) Notes: comprehensive book that mostly deals with topics that we do not touch 4

5 in this course, such as dynamics of star systems (which are collisionless) and disk dynamics. [8] Zeldovich, Ya. B. and Raizer, Yu. P., Physics of Shock Waves and High- Temperature Hydrodynamic Phenomena, (Academic Press, New York and London, 1967). Notes: A classic on shock waves. There is an economic paperback Dover edition, first published in [9] Kulsrud, R.M. Plasma Physics for Astrophysics, (Princeton, 2004). Notes: a good book on plasma physics and magnetohydrodynamics. [10] Pedlosky, J., Geophysical Fluid Dynamics, (Springer, 1987) Notes: an excellent book if you are interested in more terrestrial topics. [11] MIT Fluid Mechanics films. These are highly instructive movies available for free at Description from the website: In 1961, Ascher Shapiro [...] released a series of 39 videos and accompanying texts which revolutionized the teaching of fluid mechanics. MIT s ifluids program has made a number of the films from this series available on the web. Highly recommended! [12] Purcell, E.M. and Morin, D.J., Electricity and Magnetism, (Cambridge University Press, 2013) Notes: a nicely updated edition of Purcell s classic. Lots of solved problems. [13] Jackson, J. D., Classical electrodynamics, (Wiley, 1999) Notes: everything you always wanted to know (and more) about classical electrodynamics. 5

6 1 Hydrodynamics 1.1 Introductory remarks Fluid dynamics is one of the most central branches of astrophysics. It is essential to understand star formation, galactic dynamics (what is the origin of spiral structure?), accretion discs, supernovae explosions, cosmological flows, stellar structure (what is inside the Sun?), planet atmospheres, the interstellar medium, and the list could go on. Fluids such as water can usually be considered incompressible, because extremely high pressures of the order of thousands of atmospheres are required to achieve appreciable compressions. Air is highly compressible, but it can behave as an incompressible fluid if the flow speed is much smaller than the sound speed. Astrophysical fluids, on the other hand, must usually be treated as compressible fluids. This means that we must account for the possibility of large density changes. Astrophysical fluids usually consists of gases that are ultimately made of particles. Although they are not exactly continuous fluids, most of the time they can be treated as if they are. This approximation is valid if the mean free path of a particle is small compared to the typical length over which macroscopic quantities such as the density vary. If this is the case, one can consider fluid elements that are i) large enough that are much bigger than the mean free path and contain a vast number of atoms ii) small enough that have uniquely defined values for quantities such as density, velocity, pressure, etc. Effects such as viscosity and thermal conduction are a consequence of finite mean free paths, and extra terms can be included in the equations to take them into account in the continuous approximation. Most astrophysical fluids are magnetised. Although this can sometimes be neglected, there are many instances in which it is necessary to take explicitly into account the magnetised nature of astrophysical fluids. Therefore we shall study magnetohydrodynamics alongside hydrodynamics. 1.2 The state of a fluid In the simplest case, the state of a fluid at a certain time is fully specified by its density ρ(x) and velocity field v(x). In some cases it is necessary to know additional quantities, such as the pressure P (x), the temperature T (x), the specific entropy s(x), or in the case of magnetised fluids the magnetic field B(x). Multi-component fluids can have more than 6

7 one species defined at each point (think for example of a plasma made of positive and negative charged particles which can move at different velocities relative to one another), each with its own density and velocity, but we will not consider this case in this course. Equations of motion allow us to evolve in time the quantities that define the state of the fluid once we know them at some given time t 0. In other words, they allow us to uniquely determine ρ(x, t), v(x, t), etc at all times once their values ρ(x, t = t 0 ), v(x, t = t 0 ), etc are known for all points in space at a particular time t The continuity equation V dv ds ˆn Consider an arbitrary closed volume V that is fixed in space and bounded by a surface S (see Fig. 1). The mass of fluid contained in this volume is M(t) = ρ(x, t)dv, (1) V and its rate of change with time is (can you show why is it legit to bring the time derivative inside the integral?): dm(t) dt = V t ρ(x, t)dv. (2) Figure 1: An arbitrary volume V. Divergence theorem We can equate this with the mass that is instantaneously flowing out through the surface S. The mass flowing out the surface area element ds = ds ˆn, where ˆn is a vector normal to the surface pointing outwards, is ρv ds, thus summing contributions over the whole surface we have: dm(t) = ρv ds. (3) dt The divergence theorem states that for any vector-valued function F(x) (can you prove this?): V S dv F = S ds F(x) (4) Applying the divergence theorem with F = ρv to the RHS of Eq. (3) and then equating the result to the RHS of Eq. (2) we obtain: t ρ(x, t)dv = dv (ρv). (5) V V 7

8 Since this equation must hold for any volume V, the arguments inside the integrals must be equal at all points. 1 Hence we find: t ρ + (ρv) = 0 (6) Continuity equation This is called continuity equation and expresses the conservation of mass: what is lost inside a volume is what has outflown through the surface bounding that volume. 1.4 The Euler equation, or F = ma What is the fluid equivalent of Newton s second law F = ma? Consider a small fluid element of volume dv and mass dm = ρdv. By Newton s second law, ( ) Dv dm = sum of the forces acting on the fluid element, (7) Dt where the quantity Dv/Dt is the acceleration of the fluid element. Note that this is not the same as t v(x, t). The difference between the two is as follows: i) Dv/Dt is calculated by comparing the velocities of the same fluid element at t and t + dt, which occupies different spatial positions at different times ii) t v(x, t) is calculated by comparing velocities at the same position in space at different times. We can find the relation between the two types of derivatives, which holds for any property f(x, t) of the fluid and not just for velocities. Consider a fluid element that is initially at x(t) (see Fig. 2). Its velocity is v(x, t) and therefore after a time dt its new position will be x(t+dt) x(t) + vdt. To take the derivative following the fluid element, we must compare f at the new position at time t + dt with f at the old position at time t: Df Dt lim f (x(t + dt), t + dt) f (x(t), t) dt 0 dt (8) f (x(t) + vdt, t + dt) f (x(t), t) dt (9) tfdt + ( f) (vdt) dt (10) = t f + v f (11) 1 Suppose there is a point in which they are not equal. Then one could just integrate in the neighbourhood of that point, contradicting the result (5). Hence they must be equal. x(t) Figure 2: derivative. vdt x(t + dt) The convective 8

9 Convective derivative Thus D Dt = t + v (12) ˆn F Figure 3: A hypothetical plane slicing through the fluid. ˆn is the normal to the plane. In the general case, the force F that the fluid on one side exerts on the fluid on the other side can have any direction. In this course, we only consider cases in which ˆn and F have the same direction, and F does not depend on the orientation of the plane. D/Dt is called the Lagrangian or convective derivative, to distinguish it from the Eulerian derivative t (see also Section 1.9). Now that we have discussed the LHS of Eq. (7), let us deal with the RHS. What are the forces that act on a fluid element? The most fundamental force acting on a fluid is pressure. Consider a hypothetical plane slicing through a static (v = 0) fluid with an arbitrary orientation. What is the (vector) force that material on one side of that plane exerts on material on the other side? In the most general case the force depends on the orientation of the plane and the force itself can have any direction (different from the orientation of the plane). The thing relating the force to the orientation is in general a second-rank tensor, because it relates a vector to a vector, i.e. it is a 3 3 array of numbers. Fortunately, in this course we only consider forces that in such a static situation can always be considered isotropic, i.e. they do not depend on the orientation of ˆn (can you think of an example where this is not true?), and are directed perpendicularly to the surface at each point, i.e. they are directed along ˆn (this is usually not a good approximation in solids. Think of a twisted rubber: it is clearly not true inside it!). The force can then be quantified by a single number called pressure, which is a function of position and time, P = P (x, t). The pressure force acting on a surface area ds is P ds. A pressure exerts a net force on a fluid element only if it is not spatially uniform, otherwise the force on opposite sides cancels out. The pressure force acting on a fluid element is P dv. (13) You can show this by considering the forces on the side of a small cube of volume dv = dxdydz. If the fluid is not static, we assume that the pressure force is the same. If two layers of fluid are moving relative to each other, viscous forces can also be present. In contrast to pressure, these are not directed perpendicularly to our hypothetical plane, and will be the subject of Section Now let s put eveything together. First, substitute Eq. (12) into the LHS of (7). If the only force acting on the fluid is pressure, the RHS is 9

10 simply given by (13). After dividing by dm and using that dm = ρdv we find: t v + (v )v = P (14) ρ This is the Euler equation. This was derived assuming that pressure is the only force. Other forces can be added as needed. One of obvious importance in astrophysics is gravity. In presence of a gravitational field Φ the force per unit mass acting on a fluid element is Φ, therefore the Euler equation becomes in this case Euler equation t v + (v )v = P ρ Φ. (15) Euler equation with gravity. If the field is externally imposed then Φ is a given function of x and t; if on the contrary the field is generated by the fluid itself, it must be computed self-consistently. Other forces of astrophysical interest can arise because of the effects of magnetic fields and viscosity. These will be considered below. 1.5 The choice of the equation of state Consider the continuity equation (6) and the Euler equation (14). These two equations alone are not enough to evolve the system in time. In other words, if one is given ρ 0 (x) = ρ(x, t = t 0 ) and v 0 (x) = v(x, t = t 0 ) at t = t 0, they are not enough to determine uniquely what they are at at t > t 0. There are many possible solutions ρ(x, t), v(x, t = t 0 ) that satisfy them, and one does not know which one to choose. People say that the continuity and Euler equations do not form a complete system of differential equations, so we need one more. From the mathematical point of view, this can be understood because we have three unknowns functions (ρ, v, P ) and only two equations (6 and 14). 2 From the physical point of view, this can be understood if we consider that to find the time evolution of a fluid element we need to know the forces acting on it, but so far we have said nothing on how to determine P! It is common to relate pressure and density through an equation of state. For most astrophysical applications, it is usually a very good 2 We have five unknowns and four equations if v is considered as three scalar functions v x, v y, v z. The Euler equation is a vector equation and counts as three scalar equations. 10

11 Ideal gas approximation to consider the equation of state of an ideal gas P = ρkt µ, (16) where T is the temperature, k = JK 1 is the Boltzmann constant, µ is the mass per particle. The addition of Eq. (16) does not complete our system of equations, because we have introduced a new equation but also a new unknown, the temperature T (x, t). We have just exchanged one unknown quantity (P ) for another (T ). The following are common ways of closing the system of equations of astrophysical importance: Isothermal gas Assume that T = constant in Eq. 16. This is called an isothermal gas. In this case P is proportional to ρ. Note that the quantity kt/µ has dimensions of a velocity squared and Eq. (16) can be rewritten as P = c 2 sρ (17) where c 2 s = kt/µ = constant. (18) c s is called the sound speed, for reasons that will become clear later in the course (see Section 5.2). In general, if we follow a fluid element, it changes shape and volume during its motion. It can therefore perform work by expansion or receive work by compression from its surroundings. An adiabatic fluid is one in which this work is converted into internal energy of the fluid in a reversible manner and there are no transfers of heat or matter between a fluid element and its surroundings. The temperature of each fluid element is allowed to change, but only as a result of compression and expansion. We will see later how to consider processes able to add or subtract heat from a fluid element, such as exchanges of heat due to thermal conduction between neighbouring fluid elements, viscous dissipation, and extra heating and cooling due to radiative processes. We can find the equations governing an adiabatic gas as follows. The internal energy per unit mass of an ideal gas is Internal energy per unit mass of an ideal gas U = 11 P ρ(γ 1) (19)

12 where γ = 1 + 2/N is the adiabatic index and N is the number of degrees of freedom per particle (N = 3 for a monoatomic gas, N = 5 for a diatomic gas). Note that an isothermal gas corresponds to the limit N in which the number of degrees of freedom goes to infinity. In this case, the internal degrees of freedom act like a heat bath that keeps the temperature constant. This is also why U as γ 1. Consider again a small fluid element of volume dv, mass dm = ρdv and internal energy is dmu. As it moves through the fluid, this parcel of gas changes volume, doing work by expansion and exchanging heat with the surroundings, just like the ideal gas systems you studied in your first thermodynamics course. The first law of thermodynamics says that DU = DQ DW (20) where DU is the change in internal energy, DQ is the heat added to the system and DW is the work done by the system (all per unit mass). We use the big D to emphasise that these are changes tracked following the fluid. Differentiating (19) yields DU = (DP )/ [ρ(γ 1)] (Dρ)P/[ρ 2 (γ 1)]. The expansion work done by the fluid element due to its volume change is DW = P Dυ = (Dρ)/ρ 2, where υ 1/ρ is the specific volume, i.e. the volume per unit mass. In an adiabatic fluid, by definition, DQ = 0. Hence for an adiabatic fluid the first law becomes 1 ρ(γ 1) DP P ρ 2 (γ 1) Dρ = P Dρ (21) ρ2 rearranging and dividing by Dt we find D log (P ρ γ ) Dt = 0 (22) Adiabatic ideal fluid This last equation together with equations (6) and (14) are the equations of motion of an adiabatic gas. They are a complete system of equations that fully specifies the time evolution given the state of the fluid at t = 0. Note that this implies that the entropy per unit mass of a fluid element, which up to an unimportant additive constant is given by s = k µ(γ 1) log P ρ γ (23) 12

13 does not change in adiabatic flow, i.e. the following equation is valid: Ds = 0. (24) Dt Note also that this equation implies that s is a constant for a particular fluid element. It does not preclude different fluid elements having different values of s, it just implies that each such element will retain whatever value of s it started with. For example, a medium which is initially isothermal with a non-uniform density, such as an isothermal sphere (see Section 3.2), has s which is initially not uniform in space, hence one must be careful in not confusing t log(p ρ γ ) with D log(p ρ γ )/Dt. The particular case in which s is the same for all fluid elements is called isentropic fluid. In an isentropic fluid the pressure is related through density by P = Kρ γ where K is a constant which is the same for all fluid elements and therefore is a particular case of barotropic fluid (see below), and in this case t s = Ds/Dt = 0. But in the general case of adiabatic flow, K is not a constant and may be different for different fluid elements. An incompressible fluid is one in which fluid elements do not change in volume as they move. A liquid like water can usually be considered incompressible, because extremely high density are required to achieve significant compressions. Air is highly compressible, but it can behave as an incompressible fluid if the flow speed is much smaller than the sound speed. For an incompressible fluid the density of a fluid element is constant, Dρ Dt = 0. (25) Since the continuity equation (6) can be rewritten as Dρ Dt it implies that for an incompressible fluid + ρ v = 0, (26) Incompressible fluid v = 0 (27) Indeed, it can be shown (see problem 2) that the Lagrangian derivative of a volume element is D(dV ) Dt 13 = (dv ) v (28)

14 This means that in a time dt the volume of a fluid element changes as dv dv (1 + ( v)dt), and v is its rate of change. Hence, v = 0 means that the volume does not change, i.e. the fluid is incompressible. Equation (27) together with equations (6) and (14) form a complete system of equations. They are the equations of motion of an incompressible fluid. The density is usually a given constant, so we do not need to solve for it, while the pressure P simply assumes the value needed to sustain the flow, much like the normal reaction of a frictionless surface is as high as it needs to be to sustain an objects moving over it. Note that according to Eq. (25) the density does not need to be the same everywhere, i.e. all fluid elements, although this is often the case. An important case in astrophysics is that of a barotropic equation of state, in which the pressure is a function of density only: P = P (ρ). (29) Barotropic fluid A subclass of barotropics model often used in astrophysics is polytropic models, in which P = Kρ (n+1)/n, where K is a constant and the constant n is the polytropic index. Examples of polytropic fluids are an isothermal gas (n = ), an isentropic gas (γ = (n + 1)/n), and a degenerate gas of electrons (n = 3/2 for the non relativistic case, n = 3 for the relativistic case). 1.6 Manipulating the fluid equations When dealing with the fluid equations there are several tricks and techniques that can make your calculation faster and your life easier Writing the equations in different coordinate systems Suppose that you want to write down the continuity and Euler equations in a cylindrical or spherical coordinate system. There are several possible ways to approach this problem. You could calculate them by brute force by writing down the new variables as a function of the old ones, calculating the old derivatives as a function of the new ones using the chain rule and then substitute them in the Cartesian equations, but this is the lengthy route. A least action route is to look up on Wikipedia 14

15 the expressions of the various differential operators in your coordinate system. 3 The most instructive route, which allows you to derive most expression without the need to look them up, is probably as follows: 1. Start from expressions in vector form, which are valid in all coordinate systems. 2. Manipulate these expressions by expanding vectors using unit vectors in the coordinate system of interest. This requires knowledge of the unit vectors derivatives and how to write down the gradient operator, but these are generally easy to derive or remember. For example, let us find v in cylindrical coordinates (R, φ, z). We write = ê R R + ê 1 φ R φ + ê z (30) z and v = ê R v R + ê φ v φ + ê z v z (31) where v R is the R component of velocity, and so on. Putting these together we write the divergence as: ( ) v = ê R R + ê 1 φ R φ + ê z (ê R v R + ê φ v φ + ê z v z ) (32) z Note that the derivatives must operate on the unit vectors! In cartesian coordinates the unit vectors are constant so this does not matter, but in other coordinate systems the unit vectors generally change with position. Using the relations in Section we find: ( ) v = ê R R + ê 1 φ R φ + ê z (ê R v R + ê φ v φ + ê z v z ) (33) z = v R R + 1 v φ R φ + v z z + v R (34) R where the last term originates because of the nonvanishing derivatives of the unit vectors. As another example, let us find (v )v in cylindrical coordinates. The radial component of this vector is not (v )v R, because again we 3 coordinates 15

16 have to take the derivatives of the unit vectors. Instead we should write: [ ( )] (v )v = (ê R v R + ê φ v φ + ê z v z ) ê R R + ê 1 φ R φ + ê z (35) z (ê R v R + ê φ v φ + ê z v z ) (36) and take all the derivatives and the scalar products according to relations in Section The result should be (exercise) ( v R (v )v = v R ρ + v φ v R ρ φ + v v R z z v ) φv φ ê R (37) ρ ( v φ + v R ρ + v φ v φ ρ φ + v v φ z z + v ) φv R ê φ (38) ρ ( v z + v R ρ + v ) φ v z ρ φ + v v z z ê z (39) z Indecent indices Vector notation is nice because it is coordinate independent, thus the same expression is valid in Cartesian, Spherical or Cylindrical coordinates. However it is often useful to switch to index notation in calculations, for example in proving vector identities. The index i,j and k here will represent any of the Cartesian coordinates x, y and z. For example, v i is the ith component of v, which may be any of the three (x, y or z). The gradient operator is written i. We also use the Einstein summation convention: when an index variable appears twice in a single term it implies summation of that term over all the values of the index (unless otherwise specified). For example, the dot product between two vectors A and B is written The term v dot grad v is written A B = A i B i. (40) (v )v = (v i i )v j, (41) where the term on the right represents the component in the direction j = x, y or z. The continuity equation (6) is rewritten in index notation as follows: t ρ + i (ρv i ) = 0, (42) and the Euler equation (14): t v j + (v i i )v j = jp ρ. (43) 16

17 Note that these are three scalar equations, one for each possible value of j = x, y or z. The cross product and the curl can be written using the Levi-Civita symbol ε ijk. This is defined as +1 if (i, j, k) is (1, 2, 3), (2, 3, 1), or (3, 1, 2), ε ijk = 1 if (i, j, k) is (3, 2, 1), (1, 3, 2), or (2, 1, 3), (44) 0 if any two indices are equal to one another and Then we can write for example A = ε ijk i A j, (45) A B = ε ijk A i B j. (46) In expressions involving two cross products we have two ε ijk symbols. In those cases, the following identity is very useful: ε ijk ε imn = δ j m δ k n δ j n δ k m (47) Here the position of the index (subscript or superscript) has no meaning and different arrangements are only used for clarity. Such distinction is instead important general relativity, but we do not discuss that here. As an exercise, you can use (47) to prove the following identities: A (B C) = (A C) B (A B) C (48) ( A) = ( A) 2 A (49) (A B) = A ( B) B ( A) + (B )A (A )B (50) Tables of unit vectors and their derivatives Cylindrical unit vectors: Spherical unit vectors: ê R = (cos φ, sin φ, 0) (51) ê φ = ( sin φ, cos φ, 0) (52) ê z = (0, 0, 1) (53) ê r = (sin θ cos φ, sin θ sin φ, cos θ) = sin θê R + cos θê z (54) ê θ = (cos θ cos φ, cos θ sin φ, sin θ) = cos θê R sin θê z (55) ê φ = (sin φ, cos φ, 0) (56) 17

18 Nonvanishing derivatives of cylindrical unit vectors φ (ê R ) = ê φ (57) φ (ê φ ) = ê R (58) Nonvanishing derivatives of spherical unit vectors θ (ê r ) = ê θ (59) φ (ê r ) = sin θê φ (60) θ (ê θ ) = ê r (61) φ (ê θ ) = cos θê φ (62) φ (ê φ ) = (sin θê r + cos θê θ ) = ê R (63) More comprehensive list of properties can be found for example at: Conservation of energy A useful equation related to the conservation of kinetic energy can be obtained by taking the dot product of v with the Euler equation (14). The LHS becomes v [ t v + (v )v] = 1 2 [ t v 2 + (v )v 2] (64) = 1 Dv 2 2 Dt (65) The first step can be proved easily by switching to index notation (see Section 1.6.2). Putting this back together with the RHS we find [ 1 Dv 2 2 Dt = v P ] ρ (66) The interpretation of this equation is simple. Consider a small fluid element of mass dm. Its kinetic energy is dmv 2 /2. Then this equation simply states that the change in kinetic energy of a fluid element is the dot product between the force, dm P /ρ, and the velocity, i.e. it is the fluid mechanics equivalent of the familiar Newtonian mechanics statement that d(mv 2 /2)/dt = F v. 18

19 We can rewrite (66) in another useful way. After multiplying both sides by ρ, the LHS can be rewritten 1 2 ρdv2 Dt = 1 2 ρ ( t v 2 + v v 2) (67) = 1 2 ρ ( t v 2 + v v 2) v2 [ t ρ + (ρv)] (68) ( ) ( ) ρv 2 ρv 2 = t v (69) where in the second step we have used the continuity equation (6). The RHS can be rewritten as (v ) P = (P v) + P ( v) (70) Putting all together and rearranging we find [ ] [( ) ] ρv 2 ρv 2 t P v = P ( v) (71) We have only used the continuity and Euler equation to derive this result, so it applies both to incompressible fluids and to compressible gases. For an incompressible fluid v = 0 so the RHS side vanishes, and this constitutes a statement of the conservation of kinetic energy, 4 i.e. the total kinetic energy is a conserved quantity in an incompressible inviscid fluid. For an adiabatic gas, the term P ( v) represents the expansion work done by the gas, and kinetic energy is not a conserved quantity. To discuss the conservation of energy in this case one must consider the internal energy of the gas. The continuity equation can be rewritten (see Eq. 26) as v = D log ρ (73) Dt 4 Any statement of the type t Q + F = 0 (72) is a conservation law where Q is the conserved quantity and F is the associated flux. To see this, integrate over a finite volume and use the divergence theorem (4). This shows that the change in the amount of Qdt inside the volume is related to the V outflux quantified by F. Integrating over the whole( space and ) assuming that F = 0 at infinity, which is usually the case, you get that t QdV = 0, hence the quantity between parentheses is constant in time, i.e. it is globally conserved. What gets out of one volume just enters into an adjacent one. 19

20 From equation (22) for an adiabatic gas, D log ρ Dt = 1 γ D log P Dt (74) Eliminating D log ρ/dt from these last two equations we find v = 1 γ D log P Dt (75) = 1 P γ [ tp + (v ) P ] (76) = 1 P γ [ tp + (P v) P ( v)] (77) where in the last step we have used (70). Isolating v from this last equation we find 1 v = P (γ 1) [ tp + (P v)] (78) Substituting this into (71) and rearranging we finally find: [ ρv 2 t 2 + P ] [( ρv 2 + γ P + P ) ] v = 0 (79) γ 1 which is a statement of the conservation of energy for an adiabatic gas. The terms between the first bracket parentheses are respectively the kinetic energy per unit volume and the internal energy per unit volume. Neither mechanical nor internal energy is separately conserved, but their sum integrated over all space is conserved. The three terms between the second square brackets are the fluxes of energy due to advection of kinetic energy, to pressure forces exchanged between adjacent fluid elements, and to advection of internal energy respectively. Since we are considering an adiabatic gas, there are no other fluxes of energy related to, for example, viscous forces or thermal conduction. The sum of kinetic plus internal energy must be conserved even if the gas is viscous, since dissipation does not constitute an external heat source (see Section 1.13). What about energy conservation in an isothermal fluid? Well, energy is not conserved in an isothermal fluid. 5 One must imagine that any excess/lack of heat is removed/provided as needed by an external heat bath to each fluid element, in such a way that the temperature is constant Conservation of energy for an adiabatic fluid. 5 If one tries to see an isothermal gas as the γ limit as discussed in Sect. 1.5, Eq. (71) diverges so it cannot be applied! 20

21 during its motion. In an astrophysical context, it is sometimes a useful approximation to treat the interstellar medium as an isothermal fluid, in which the thermal balance is maintained for example by external heating sources and collisional cooling. 1.8 Conservation of momentum Conservation of momentum can be proved starting from the continuity (6) and Euler (14) equations. Multiplying (14) by ρ and considering the ith component we can rewrite the LHS as Conservation of momentum. ρ Dv i Dt = ρ [ tv i + (v ) v i ] (80) = ρ [ t v i + (v ) v i ] + v i [ t ρ + (ρv)] (81) = t (ρv i ) + [(ρv i )v] (82) = t (ρv i ) + j [(ρv i )v j ] (83) where in the second step we have added a term which vanishes thanks to the continuity equation. The RHS can be rewritten as Putting all together and rearranging we find i P = δ ij j P (84) t (ρv i ) + j [(ρv i )v j + δ ij P ] = 0 (85) This equation is valid both for an incompressible and compressible fluid and is a statement of the conservation of momentum. To see this, just integrate over all space. If all the fluid quantities v i, P and ρ vanish at infinity, then the integral of the second term in Eq. (85) disappears and we are left with ( ) t dv ρv i = 0 (86) V which means that the quantity between parentheses, i.e. the total momentum in the ith direction, is constant in time. 1.9 Lagrangian vs Eulerian view In astrophysics you often hear talking about Lagrangian view as opposed to Eulerian view. What is meant exactly may depend on who you are speaking to. In practice, most of the time it is referred to equivalent ways of solving the equations of hydrodynamic. Since they are equivalent, one should choose the most convenient for the problem at hand. 21

22 In a Eulerian description, quantities are written as a function of fixed spatial coordinates. For example, ρ = ρ(x, t) is the density at a fixed location in space x. If we write the continuity equation as t ρ(x, t) + (ρ(x, t)v(x, t)), it is meant that t takes the time derivatives by comparing quantities at different times but at the same location in space, and are derivatives obtained comparing quantities at neighbouring locations in space at the same time. In a Lagrangian description, quantities are written as a function of coordinates that move with the flow. For example, suppose that we label all fluid elements by the position x 0 they had ad time t = t 0. Then we can write ρ = ρ(x 0, t) meaning what is the density at time t > t 0 of the fluid element that was at x = x 0 at time t = t 0? In general the position of such a fluid element changes with time, so ρ(x 0, t) is not the density of a fixed point in space. It is the density that we would see by following a given fluid element in time. Since the convective derivative (12) follows the flow, it is also called the Lagrangian derivative. Equations of motion written in terms of D/Dt are sometimes said to be in Lagrangian form Vorticity In fluid dynamics it often useful to consider the vorticity ω = v. (87) Definition of vorticity. The physical meaning of this quantity can be obtained considering two short fluid line elements which are perpendicular at a certain instant and move with the fluid (see Fig. 4). The vorticity is twice the average angular velocity of two such short fluid elements. In this sense, the vorticity is a measure of the local degree of spin, or rotation, of the fluid. Note that this may not always be the same of our intuitive notion of rotation; if the two lines rotate in opposite directions with equal angular velocity the vorticity is zero. Note also that ω is not directly related to global rotation. Velocity configurations are possible for which there is no global rotation but ω 0. As an exercise, consider for example the vorticity of the flow fields defined by v = βyê x, where β is a constant, and v = (k/r)ê φ. 22 A D C vx y v y x B Figure 4: Physical meaning of vorticity. ω is taken to point out of the page. The angular velocity of the two perpendicular fluid element lines AB and CD is x v y and y v x respectively. Hence their average angular velocity is ( x v y y v x )/2 = ω/2.

23 The vorticity equation To obtain an equation for the evolution of vorticity, consider the following identity: v ( v) = 1 2 v2 (v ) v (88) which can be proved using (70). equation (14) as Using this we can rewrite the Euler t v v ω v2 = P ρ (89) Taking the curl of this equation and using that the curl of a gradient vanishes, we obtain t ω (v ω) = 1 ( ρ P ). (90) ρ2 Using identity (50) with A = v and B = ω, and remembering that the divergence of the curl vanishes we can rewrite it as: t ω + (v ) ω = (ω ) v ω ( v) + 1 ( ρ P ). (91) ρ2 Now substitute v from the continuity equation in the form (73): Dω Dt = (ω ) v + ω D log ρ Dt + 1 ( ρ P ). (92) ρ2 Vorticity equation. which can be rewritten as ( ) [( ) ] D ω ω = v + 1 ( ρ P ) Dt ρ ρ ρ2 (93) Barotropic condition. This is the vorticity equation, which describes the evolution of vorticity. The quantity ω/ρ is called potential vorticity. Note that in the presence of a gravitational field the vorticity equation would be exactly the same since the extra term vanishes in the step in which we take the curl of the Euler equation (the curl of a gradient is zero, and the gravitational force is the gradient of Φ). Now assume that ρ P = 0. (94) This is called barotropic condition and it means that surfaces of constant density are the same as surfaces of constant pressure. It is satisfied 23

24 for barotropic fluids (29), which include isothermal and isentropic fluids as particular cases, and for incompressible fluids (27), for which the density is constant. Under this assumption the vorticity equation becomes ( ) [( ) ] D ω ω = v. (95) Dt ρ ρ If the flow is two-dimensional, so that ω is always perpendicular to v, the term on the right hand side vanishes which implies that the potential vorticity is conserved, i.e. it is constant for a given fluid element. In the case of an incompressible flow, we can simplify ρ and the vorticity itself is conserved. These are powerful constraints: suppose for example that we know that the flow is steady, so that it does not change with time, and consider a streamline. If ω/ρ or ω is zero at some point along the streamline, it will be zero at all points along the streamline. Often we know that the potential vorticity (or the vorticity) vanishes at special points, for example at infinity, and we can use this condition to conclude that it is zero at all points reached by streamlines, which could be everywhere. In the presence of viscosity (Section 1.13) the vorticity is not conserved anymore, even for an incompressible fluid. In this case it is possible to study how the vorticity is diffused, but we will not do it in this course (if you are interested, see for example [1]) Kelvin circulation theorem Consider a closed curve that moves with the fluid, see Fig. 5. The integral of the component of the velocity parallel to the curve around the closed curve C = v dl (96) γ is called the circulation. Kelvin circulation theorem states that, in an inviscid fluid in which the barotropic condition (94) is valid, this integral is constant in time if we follow the fluid, i.e. C(t 1 ) = C(t 2 ). To prove Kelvin s theorem, let us consider a slightly more general situation, which will be useful later when we consider flux freezing in magnetohydrodynamics. Suppose you have an equation of the following form t A = v ( A) + f (97) where f is a generic function. We will now prove that Γ(t) = A dl, (98) γ C(t 1 ) C(t 2 ) Figure 5: Kelvin circulation theorem. 24

25 Stokes theorem. calculated following the fluid, is constant in time, i.e. Γ(t 1 ) = Γ(t 2 ). First, take the Lagrangian derivative of Γ(t): 6 ( ) ( ) DΓ(t) D D = Dt Dt A dl + A Dt dl (99) γ We need to show that this is zero. (97) can be rewritten with the help of identity (47) as: DA i = v j i A j + i f (100) Dt Using this equation the first term on the RHS of (99) can be rewritten as: ( ) D Dt A dl = (v j i A j + i f) dl i (101) γ Next, use the result of problem 2, equation (150) γ γ D dl = (dl ) v (102) Dt to rewrite the second term on the RHS of (99) as: ( ) D A Dt dl = A j dl i i v j (103) Adding (101) and (103) gives D (A dl) = dl i i (A j v j + f) = Dt γ γ γ γ γ dl (A v + f) = 0 (104) which vanishes because the integral of a gradient over a closed loop is zero, which proves the theorem. Note that using Stokes theorem, which states that A dl = ( A) ds (105) γ S where S is an open surface 7 bounded by the curve γ, we can rewrite (98) as Γ(t) = ( A) ds. (106) S 6 as an exercise, you can show that it is allowed to take the derivative inside the integral. 7 Note that this need not to be flat, it can be any of the infinite possible surfaces that have γ as boundary. This implies that the integral of a curl over a closed surface is zero. 25

26 This can be interpreted as if the vector A is frozen into the fluid. A small patch of fluid will retain during its motion the amount of flux of this vector that was given initially. To see how this theorem applies with A = v and recover Kelvin theorem, consider the Euler equation in its form (89): t v v ω v2 = P ρ (107) To bring this equation in the form (97), note that if the barotropic condition (94) holds, then our fluid is either barotropic or incompressible. In both cases we can define a function H such that P ρ For a barotropic fluid this is H(ρ) = while for an incompressible fluid it is = H (108) dρ P (ρ) ρ (109) H(ρ) = P ρ (110) hence, (107) is of the form (97) with f = H v 2 /2 and A = v. The vorticity ω is frozen into the fluid Steady flow: the Bernoulli s equation A flow is said to be steady when all quantities do not depend on time ( t = 0). Of course, by this we do not mean that the fluid does not move, but only that the velocity and all other quantities do not change with time at each fixed point in space. A useful theorem for this type of flow can be derived as follows. Consider Euler equation in the presence of a gravitational field (15): t v + (v )v = P ρ Φ (111) Then assume that the flow is steady so that t v = 0 and then use the identity (88) to rewrite this equation as: 1 2 v2 v ( v) = P ρ 26 Φ (112)

27 Bernoulli s theorem. Euler equation in a rotating frame. Now assume that the flow is either barotropic, P = P (ρ), or incompressible, so that the function H given by (109) or (110) can be defined. Then the Euler equation becomes 1 2 v2 v ( v) = H Φ (113) The term v ( v) is perpendicular to v at each point in the flow. Hence, if we take the scalar product of with v, we find that ( ) 1 v 2 v2 + H + Φ = 0 (114) Since the direction of v is the same as the tangent to streamlines, this means that the quantity 1 2 v2 + H + Φ = constant (115) is constant along each streamline. This is known as Bernoulli s theorem. It is valid if the flow is steady, barotropic or incompressible, and nonviscous. For an incompressible fluid, H = P/ρ, and the theorem says that where velocity goes up, pressure goes down. Note that this theorem says nothing about the constant being the same on different streamlines, only that it remains constant along each one. The constant is the same if we make one further assumption, namely that the flow is irrotational, which means that v = 0. (116) In this case, we do not need to take the scalar product with v in the derivation above, and the constant is the same throughout the whole fluid Rotating frames In astrophysics it is often useful to work in a frame rotating with constant angular speed Ω. This may be the frame in which a binary system is stationary, a rotating fluid is at rest, or a spiral pattern is stationary. In a rotating frame, we simply add the Coriolis and Centrifugal forces to the RHS of the Euler equation (14), which gives the forces acting on fluid elements, just as we would do for Newtonian point-particle mechanics. Thus the Euler equation in a rotating frame is: t v + (v )v = P ρ 2Ω v Ω (Ω x) 27 (117)

28 where 2Ω v is the Coriolis force, Ω (Ω x) is the centrifugal force and velocities are measured in the rotating frame and are related to those in the inertial frame by v inertial = v + Ω x. (118) Note that if Ω = Ωê z we can rewrite the centrifugal force as Ω (Ω x) = RΩ 2 ê R Viscosity and thermal conduction The fact that the mean free path is small but finite has the consequence that particles can be exchanged between adjacent fluid elements, which creates transfers of momentum and energy in addition to those that we have studied in the previous sections. This is the origin of dissipative processes such as viscosity and thermal conduction (see Fig. 6). Viscosity creates forces that tend to prevent velocity gradients, i.e. it is a force that opposes layers moving relative to each other. These are not contained in the equations considered so far. To obtain the equation of motion of a viscous fluid we need to modify our equations. The continuity equation (6) is unchanged. We need to add other terms to the Euler equation (14). To do this, it is convenient to start from the version written as conservation of momentum of an inviscid fluid, equation (85): t (ρv i ) + j [Π ij ] = 0 (119) In this equation, the term Π ij = ρv i v j + δ ij P (120) represents the flux of the ith component of momentum in the j direction. The equations of motion for a viscous flow are obtained adding to Π ij an additional term σ ij that accounts for the transfer of momentum due to viscous processes: Π ij = ρv i v j + δ ij P + σ ij. (121) The form of σ ij is derived heuristically, i.e. it is not a derivation from first principles, much like the description of friction in first year mechanics. We do not go through the whole derivation here, and the reader is referred for example to [2] or [1]. The derivation proceeds along the following lines. σ ij is required to satisfy the following properties: 1. σ ij = 0 when there are no relative motions between different parts of the fluid. This means that σ ij must depend only on the space derivatives of the velocity. A l S l B Figure 6: Particles exchanged between adjacent fluid elements A and B are the origin of viscosity. Molecules are exchanged between layers whose thickness is of the order of the mean free path l. Fluid elements are much bigger than l. If A and B are moving in the y direction with different velocities, particles escaping from the faster fluid element will accelerate the slower one and viceversa. x 28

29 2. We assume that σ ij depends only on the first derivatives of the velocity and that it is a linear function of these first derivatives. This is a heuristic reasonable approximation. It is true for example if an expression for the momentum transfer is obtained from the simple picture sketched in Fig We require that σ ij does not look special in any inertial frame of reference. If this weren t the case, that frame would be different from the other frames and Galilean invariance would not be satisfied. Thus we require that when we change frame of reference (i.e. we perform a translation, rotation or add a constant relative motion) σ ij transforms as a second-rank tensor We require σ ij to vanish when the whole fluid is in rigid rotation since in such a motion no internal friction occurs, i.e. σ ij = 0 if v = Ω x where Ω is a constant vector. 5. We require σ ij to vanish for uniform expansion. This is not a necessary requirement and there is a type of friction called second viscosity, which we do not consider in this course, which arises from such a term. This is important for example when internal degrees of freedom are slow in being excited, so that a fast expansion can result in instantaneous thermodynamic equilibrium not being satisfied (see for example [2] and [8] for more information), but will not be important in this course. Viscous stress tensor The most general second rank tensor satisfying the above conditions is σ ij = η ( j v i + i v j 23 ) δ ij k v k (122) where η is a parameter called dynamic viscosity and in general may be a function of fluid quantities such as density and pressure, η = η(ρ, P,... ). Its dimensions are [η] = mass length time. (123) Other properties of the viscous stress tensor worth mentioning are that it is symmetric, σ ij = σ ji, and that it has zero trace, σ ii = 0, which is 8 For example, the expression σ ij = v 1 cannot be true in all inertial frames. Hence the frames in which it is true would be special. The expression σ ij = (i+1) v j is also not allowed, while the expression σ ij = i v j is allowed because it has the same form in all inertial frames. 29

30 a consequence of condition (5) above (the trace of σ ij is proportional to v, which is associated with expansions). Substituting (121) where σ ij is given by (122) back into (119) and performing a few steps 9 one arrives at: t v i + v j j v i = ip ρ + 1 ρ j (σ ij ) (124) This equation replaces the Euler equation for a fluid in which viscous processes occur. If we further assume that η = constant so that we can take it out of the derivatives we obtain the Navier-Stokes equation 10 t v + ( v) v = P ρ η ρ 2 v + η [ ( v)] Navier-Stokes equation 3ρ (125) where 2 v = j j v i and ( v) = i ( j v j ). We have seen in Section 1.5 that the continuity and Euler equations alone are not enough to form a complete system of equations that determine the evolution in time of the fluid. Analogously, the continuity equation plus the Navier-Stokes equation are not a complete system of equations. Among the possible ways to complete the equations discussed in Section 1.5, the incompressible approximation and the isothermal one remain valid possibilities. 11 However, the adiabatic approximation cannot be valid anymore because viscous processes will dissipate energy which according to the first law will be converted into internal energy. For a viscous fluid the conservation of entropy equation (23) (or equivalently 22) cannot hold anymore. To see how these should be replaced, we need to find out what is the dissipation due to viscous processes and then use the first law of thermodynamics to equate them with the increase in the internal energy of fluid elements. Dissipation means loss of mechanical energy. For a viscous fluid the change in mechanical energy was expressed by (71). We can derive an analogous equation for the case of a viscous fluid. Taking the dot product between v and (124) and after manipulations similar to the non viscous 9 You will need to use the continuity equation (6). 10 Sometimes what is called the Navier-Stokes equation is (125) for an incompressible fluid, i.e. assuming v = 0, while some other times it is (125) including an additional term that accounts for the second viscosity briefly mentioned above. 11 In the isothermal case one must assume that the processes that keep the temperature constant are faster than any other process that may change the temperature of fluid elements, such as viscous dissipation. 30